Question

Prove that limit does not exist using epsilon delta proof : `lim_ (x->0) |x|/x` ? Please show the scratch work too.

Answer 1

For any `x > 0` we have `absx/x = 1` while for any `x < 0` we have `absx/x = -1`.

We therefore have:

`lim_(x->0^+) absx/x = 1`

`lim_(x->0^-) absx/x = -1`

and as the limits frm the right and from the left do not coincide, the limit cannot exist.

To see it in terms of `epsilon-delta` we can use Cauchy' necessary condition, stating that:

`lim_(x->0) f(x)`

exists and is finite only if given a number `epsilon` we can find a `delta_epsilon` such that:

`(1) " " abs(f(x_1)-f(x_2)) < epsilon`

for any `x_1,x_2 in (-delta_epsilon, delta_epsilon)`.

However, if we take `epsilon = 1`, then for any `delta` we can choose `x_1 in (0,delta)` and `x_2 in (-delta,0)`, so that both points belong to `(-delta, delta)` but:

`abs ( abs(x_1)/x_1 -abs(x_2)/x_2) = abs( 1-(-1)) = 2 > 1`

So that no `delta` can be found to satisfy condition `(1)`.

On the other hand as the function is bounded the limit cannot be `+-oo` either.

So we can conclude that the limit does not exists.